91Ó°ÊÓ

Probability is the likelihood that an event will occur and is calculated by dividing the number of favourable outcomes by the total number of possible outcomes. The simplest example is a coin flip. When you flip a coin there are only two possible outcomes, the result is either heads or tails.

\({\bf{X}}\)is used to denote a random variable.

\(X = \)The number of functional components in a system made up of $n$ separate components.

It is self-evident that the random variable in question follows the Binomial Distribution. The criteria for such a distribution in a \(3\)-out-of-\(5\) system would be

\(n = 5\)

\(p = 0.9,{\rm{ which is given in the exercise}}{\rm{. }}\)

The cdf of the binomial random variable \({\rm{X}}\)with parameters \({\rm{n}}\) and \({\rm{p}}\)is the Cumulative Density Function.

\(B(x;n,p) = P(X \le x) = \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n.\)

Theorem:

\(b(x;n,p) = \left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}n\\x\end{array}} \right){p^x}{{(1 - p)}^{n - x}}}&{,x = 0,1,2, \ldots ,n}\\0&{,{\rm{ otherwise }}}\end{array}} \right.\)

The system work when \(X \ge 3\), therefore

\(P(X \ge 3)\)\( = 1 - P(X \le 2)\)

\( = 1 - B(2;5,0.9)\)

\( = 1 - 0.009\)

\( = 0.991\)

Hence, the probability that a 3-out-of-5 system functions is \(P(X \ge 3) = 0.991\).